Title says it all. I'm guessing it shouldn't be possible for an estimator to be UMVUE and yet not consistent. Is there a counter-example to this? Or is there a proof that it can't happen?


1 Answer 1


One counterexample is when there's no consistent estimator. Suppose $X_i\sim N(\mu_i,1)$ where $\mu_i$ are all distinct unknown parameters. The UMVUE of any $\mu_i$ is $X_i$, but it's not consistent.

You can make matters worse. Suppose you a have single $X\sim N(\theta,1)$ together with $Y_i\sim\text{Bernoulli}(p)$ with $\mathrm{logit}\,p=\theta$ for $i=1,\dots,n$. Then $\mathrm{logit}\, \bar Y_n$ is a consistent estimator for $\theta$ as $n\to\infty$, but no function of $Y_i$ is unbiased for $\theta$. This means the UMVUE of $\theta$ is just $X$, which is not consistent.

However, for iid $X_i$, $i=1,\dots,n$, if we assume there exists some unbiased estimator $\tilde\theta$ that has finite variance for all $n$ greater than some $n_0$, the MVUE must be consistent. We can divide the $n$ observations into $[n/n_0]$ blocks of size at least $n_0$ and take the average of $\tilde\theta$ for each block. If $\sigma^2$ is the variance of each block-specific $\tilde\theta$ then the variance of the average is $\sigma^2/[n/n_0]$, which goes to zero as $n\to\infty$. So there is a consistent unbiased estimator and so the MVUE is also consistent.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.